Was the Bavarian Abitur too hard this time?


Bavaria is known for its famous Oktoberfest… and within Germany also for its presumably difficult Abitur, a qualification granted by university-preparatory schools in Germany.

A mandatory part for all students is maths. This year many students protested that the maths part was way too hard, they even started an online petition with more than seventy thousand supporters at this time of writing!

It is not clear yet whether their marks will be adjusted upwards, the ministry of education is investigating the case. As a professor in Bavaria who also teaches statistics I will take the opportunity to share with you an actual question from the original examination with solution, so read on…

Let us have a look at the first (and easiest) question in the stochastics part:

Every sixth visitor to the Oktoberfest wears a gingerbread heart around his neck. During the festival 25 visitors are chosen at random. Determine the probability that at most one of the selected visitors will have a gingerbread heart.

Before you read on try to solve the task yourself…

Of course students are not allowed to use R in the examination but in general the good thing about this kind of questions is that if you have no idea how to solve them analytically solving them by simulation is often much easier:

set.seed(12)
N <- 1e7
v <- matrix(sample(c(0L, 1L), size = 25 * N, replace = TRUE, prob = c(5/6, 1/6)), ncol = 25)
sum(rowSums(v) <= 1) / N
## [1] 0.062936

The answer is about 6.3\%.

Now for the analytical solution: “At least one” implies that we have to differentiate between two cases, “no gingerbread heart” and “exactly one gingerbread heart”. “No gingerbread heart” is just \left(\frac{5}{6}\right)^{25}. “Exactly one gingerbread heart” is 25\cdot\frac{1}{6}\cdot\left(\frac{5}{6}\right)^{24} because there are 25 possibilities of where the gingerbread heart could occur. We have to add both probabilities:

(5/6)^25 + 25*1/6*(5/6)^24
## [1] 0.06289558

If you know a little bit about probability distributions you will recognize the above as the binomial distribution:

pbinom(q = 1, size = 25, prob = 1/6)
## [1] 0.06289558

Of course it is a little unfair to judge just on basis of the easiest task and without knowing the general maths level that is required. But still, I would like to hear your opinion on this. Also outsiders’ views from different countries and different educational systems are very welcome! So, what do you think:

Was the Bavarian Abitur too hard this time? Please leave your reply in the comment section below!

Update (June 7, 2019): The Bavarian Ministry of Education has decided now that the marks will not be subsequently adjusted.

6 thoughts on “Was the Bavarian Abitur too hard this time?”

  1. Ha! Answering this question in R is easy – if you can figure out the binomial case. But… I would be totally screwed if I had to use a calculator for that answer, or even do it by hand. I am sure the pupils are well trained for such a worse case scenario, but scientists and R-users might be not.

  2. Hello, in my opinion, the analytical solution is trivial to any not-so-bad high school pupil, or at least it was when I was myself in high school (and I was pretty bad back then). If the abitur is as hard as you say, it fully belongs in it according to this criterion.
    Asking the probability of having exactly n people wearing a gingerbread heart (2 < n < 23) would be a bit harder since it would at least require the student to know the formula.
    The numeric calculation, on the other hand, is the hard part, not because it is hard, but because it is error-prone. It is also usually viewed as the least interesting because you can most often use a computer to do it in your place.

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